TITLE: Even bi-coloring of regular graphs
QUESTION [7 upvotes]: I am interested in exhaustively rendering quartic (4-regular) simple graphs of increasing order and bi-colored black and white such that the closed neighborhood of each vertex contains an even number of black vertices. Here, the closed neighborhood of a vertex is defined as the vertex itself plus its adjacent vertices. 
A non-trivial (not 'all white') example is provided by the quartic graph of order five ($K_5$, the complete graph with 5 vertices) with two or four vertices colored black and the remaining ones colored white. 
How to generate non-trivial graphs of higher order? Can this be done exhaustively?

REPLY [3 votes]: Not really a solution, but I think an important insight.
If we view 2-colorings as a maps from the vertex set to $\mathbb{Z}/2\mathbb{Z}$ sending black vertices to 1 and white vertices to 0, then the space of all 2 colorings on a fixed graph forms a vector space by pointwise addition.  The condition that a neighborhood has an even number of black vertices translates to a homogeneous linear equation, so in particular the space of good colorings is a vector subspace. Hence there will always be a power of 2 number of such good colorings.
Let $A$ be the adjacency matrix of the graph, then these linear equations are just of the form $(A+I)x=0$ where I is the identity matrix and $x$ is a vector indicating which vertices we give which color. So the graph has a nontrivial good coloring iff the matrix $A + I$ doesn't have full rank over $\mathbb{Z}/2\mathbb{Z}$ which just means that $A+I$ has even determinant. As far as describing such graphs goes, I'm at a loss.